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jasenka [17]
3 years ago
11

How to find percentage abundanace

Chemistry
1 answer:
Ghella [55]3 years ago
5 0

Answer and Explanation:

How to Calculate the Percent Abundance of an Isotope

Step 1: Find the Average Atomic Mass. Identify the atomic mass of the element from your isotopic abundance problem on the periodic table. ...

Step 2: Set Up the Relative Abundance Problem. ...

Step 3: Solve for x to Get the Relative Abundance of the Unknown Isotope. ...        

 Step 4: Find percent abundance.

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Metals posses varying reduction potentials. Iron can reduce Cu+2 (aq) to copper metal. Silver can reduce Au+ (aq) to gold metal.
natta225 [31]

<u>Answer:</u> The true statement is iron can reduce Au^+(aq) to gold metal

<u>Explanation:</u>

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

A+BC\rightarrow AC+B

Metal A is more reactive than metal B.

We are given:

Iron can reduce copper, silver can reduce gold, sodium can reduce iron and copper can reduce silver metal.

The increasing order of reactivity thus follows:

Au

where, sodium is most reactive and gold is least reactive

For the given options:

<u>Option 1:</u>  Copper cannot easily reduce sodium ion to sodium metal because it is less reactive.

Cu(s)+Na^+(aq.)\rightarrow \text{ No reaction}

<u>Option 2:</u>  Iron cant easily reduce gold ion to gold metal because it is more reactive.

Fe(s)+3Au^+(aq.)\rightarrow Fe^{3+}(aq.)+3Au(s)

<u>Option 3:</u>  Silver cannot easily reduce iron ion to iron metal because it is less reactive.

Ag(s)+Fe^{3+}(aq.)\rightarrow \text{ No reaction}

Hence, the true statement is iron can reduce Au^+(aq) to gold metal

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3 years ago
The highly reactive elements in group 7a are known for forming salts.what are they called
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Group 7A are halogens 

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Forecasting the weather for a specific area is relatively easy because of all the technology that is available.
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Answer:

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For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the
AlladinOne [14]

Answer:

<h3>1. 10 e⁻</h3>

Oxidation numbers

I₂O₅(s): I (5+); O(2-)

CO(g): C(2+); O(2-)

I₂(s): I(0)

CO₂(g): C(4+); O(2-)

<h3>2. 4 e⁻</h3>

Oxidation numbers

Hg²⁺(aq): Hg(2+)

N₂H₄(aq): N(2-); H(1+)

Hg(l): Hg(0)

N₂(g): N(0)

H⁺(aq): H(1+)

<h3>3. 6 e⁻</h3>

Oxidation numbers

H₂S(aq): H(1+); S(2-)

H⁺(aq): H(1+)

NO₃⁻(aq): N(5+); O(2-)

S(s): S(0)

NO(g): N(2+); O(2-)

H₂O(l): H(1+); O(2-)

Explanation:

In order to state the total number of electrons transferred we have to identify both half-reactions for each redox reaction.

1.  I₂O₅(s) + 5 CO(g) → I₂(s) + 5 CO₂(g)

Oxidation: 10 e⁻ + 10 H⁺(aq) + I₂O₅(s) → I₂(s) + 5 H₂O(l)

Reduction: 5 H₂O(l) + 5 CO(g) → 5 CO₂(g) + 10 H⁺(aq) + 10 e⁻

2. 2 Hg²⁺(aq) + N₂H₄(aq) → 2 Hg(l) + N₂(g) + 4 H⁺(aq)

Oxidation: N₂H₄(aq) → N₂(g) + 4 H⁺(aq) + 4 e⁻

Reduction: 2 Hg²⁺(aq) + 4 e⁻ → 2 Hg(l)

3. 3 H₂S(aq) + 2H⁺(aq) + 2 NO₃⁻(aq) → 3 S(s) + 2 NO(g) + 4H₂O(l)

Oxidation: 3 H₂S(aq) → 3 S(s) + 6 H⁺(aq) + 6 e⁻

Reduction: 8 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O

5 0
3 years ago
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